1. Chapter 6: Oxidation-Reduction Reactions
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop

2. Oxidation-Reduction Reactions
Electron transfer reactions
Electrons transferred from one substance to another
Originally only combustion of fuels or reactions of metal with oxygen
Important class of chemical reactions that occur in all areas of chemistry & biology
Also called redox reactions

3. Oxidation–Reduction Reactions
Involves 2 processes:
Oxidation = Loss of Electrons (LEO)
Na  Na+ + e Oxidation Half-Reaction
Reduction = Gain of electrons (GER)
Cl2 + 2e  2Cl Reduction Half-Reaction
Net reaction:
2Na + Cl2  2Na+ + 2Cl
Oxidation & reduction always occur together
Can't have one without the other

4. Oxidation Reduction Reaction
Oxidizing Agent
Substance that accepts e's
Accepts e's from another substance
Substance that is reduced
Cl2 + 2e  2Cl–
Reducing Agent
Substance that donates e's
Releases e's to another substance
Substance that is oxidized
Na  Na+ + e–

5. Redox Reactions Very common Chlorine Bleach
Batteries—car, flashlight, cell phone, computer
Metabolism of food
Combustion
Chlorine Bleach
Dilute NaOCl solution
Cleans through redox
reaction
Oxidizing agent
Destroys stains by oxidizing them

6. Redox Reactions Ex. Fireworks displays Net: 2Mg + O2  2MgO
Oxidation:
Mg  Mg2+ + 2e
Loses electrons = Oxidized
Reducing agent
Reduction:
O2 + 4e  2O2
Gains electrons = Reduced
Oxidizing agent

7. Your Turn!
Which species functions as the oxidizing agent in the following oxidation-reduction reaction?
Zn(s) Pt2+(aq)  Pt(s) + Zn2+(aq)
Pt(s)
Zn2+(aq)
Pt2+(aq)
Zn(s)
None of these, as this is not a redox reaction.

8. Guidelines For Redox Reactions
Oxidation & reduction always occur simultaneously
Total number of electrons lost by one substance = total number of electrons gained by second substance
For a redox reaction to occur, something must accept electrons that are lost by another substance

9. Oxidation Numbers Bookkeeping Method Way to keep track of electrons
Not all redox reactions contain O2 & give ions
Covalent molecules & ions often involved
Ex. CH4, SO2, MnO4–, etc.
Defined by set of rules
How to divide up shared electrons in compounds with covalent bonds
Change in oxidation number of element during reaction indicates redox reaction has occurred

10. Hierarchy of Rules for Assigning Oxidation Numbers
Oxidation numbers must add up to charge on molecule, formula unit or ion.
Atoms of free elements have oxidation numbers of zero.
Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively.
H & F in compounds have +1 & –1 oxidation numbers, respectively.
Oxygen has –2 oxidation number.
Group 7A elements have –1 oxidation number.

11. Hierarchy of Rules for Assigning Oxidation Numbers
Group 6A elements have –2 oxidation number.
Group 5A elements have –3 oxidation number.
When there is a conflict between 2 of these rules or ambiguity in assigning an oxidation number, apply rule with lower oxidation number & ignore conflicting rule.
Oxidation State
Used interchangeably with oxidation number
Indicates charge on monatomic ions
Iron (III) means +3 oxidation state of Fe or Fe3+

12. Ex. Assigning Oxidation Number
Li2O
Li (2 atoms) × (+1) = (Rule 3)
O (1 atom) × (–2) = –2 (Rule 5)
sum = (Rule 1)
+2 –2 = 0 so the charges are balanced to zero
CO2
C (1 atom) × (x) = x
O (2 atoms) × (–2) = –4 (Rule 5)
sum = (Rule 1)
x  4 = 0 or x = +4
C is in +4 oxidation state

13. Learning Check Assign oxidation numbers to all atoms: Ex. ClO4
O (4 atoms) × (–2) = –8
Cl (1 atom) × (–1) = –1
(molecular ion) sum ≠ –1 (violates Rule 1)
Rule 5 for O comes before Rule 6 for halogens
Cl (1 atom) × (x) = x
sum = –1 (Rule 1)
–8 + x = –1 or x = 8 –1
So x = +7; Cl is oxidation state +7

14. Learning Check Assign Oxidation States To All Atoms: MgCr2O7 KMnO4
Mg =+2; O = –2; and Cr = x (unknown)
+2 + 2x + {7 × (–2)} = 0
2x – 12 = 0 x = +3
Cr is oxidation # of +3
KMnO4
K =+1; O = – 2; so Mn = x
+1 + x + {4 × (–2)} = 0
x – 7 = 0 x = +7
Mn is oxidation # of +7

15. Your Turn! What is the oxidation number of each atom in H3PO4?
A. H = –1; P = +5; O = –2
B. H = 0; P = +3; O = –2
C. H = +1; P = +7; O = –2
D. H = +1; P = +1; O = –1
E. H = +1; P = +5; O = –2

16. Redefine Oxidation-Reduction in Terms of Oxidation Number
A redox reaction occurs when there is a change in oxidation number.
Oxidation
Increase in oxidation number
e loss
Reduction
Decrease in oxidation number
e gain

17. Using Oxidation Numbers to Recognize Redox Reactions
Sometimes literal electron transfer:
Cu: oxidation number decreases by 2  reduction
Zn: oxidation number increases by 2  oxidation

18. Using Oxidation Numbers to Recognize Redox Reactions
Sometimes electron transferred in "formal" sense.
O: oxidation number decreases by 2  reduction
C: oxidation number increases by 8  oxidation

19. Ion Electron Method Way to balance redox equations
Must balance both mass & charge
Write skeleton equation
Only ions & molecules involved in reaction
Break into 2 half-reactions
Oxidation
Reduction
Balance each half-reaction separately
Recombine to get balanced net ionic equation

20. Balancing Redox Reactions
Some Redox reactions are simple:
Ex. 1 Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
Break into half-reactions
Zn(s)  Zn2+(aq) + 2e oxidation
 LEO
Reducing agent
Cu2+(aq) + 2e  Cu(s) reduction
 GER
Oxidizing agent

21. Example 1 Zn(s)  Zn2+(aq) + 2e oxidation
Cu2+(aq) + 2e  Cu(s) reduction
Each half-reaction is balanced for atoms
Same # atoms of each type on each side
Each half-reaction is balanced for charge
Same sum of charges on each side
Add both equations algebraically, canceling e’s
NEVER have e's in net ionic equation
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)

22. Balancing Redox Equations in Aqueous Solutions
Many redox reactions in aqueous solution involve H2O and H+ or OH
Balancing the equation cannot be done by inspection.
Need method to balance equation correctly
Start with acidic solution then work to basic conditions

23. Redox in Aqueous Solution
Ex. 2 Mix solutions of K2Cr2O7 & FeSO4
Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+
Cr2O72– is reduced to form Cr3+
Acidity of mixture decreases as H+ reacts with oxygen to form water
Skeletal Eqn. Cr2O72– + Fe2+  Cr Fe3+
Ox. # Cr = Fe = Cr = Fe = +3

24. Ion-Electron Method – Acidic Solution
1. Divide equation into 2 half-reactions
2. Balance atoms other than H & O
3. Balance O by adding H2O to side that needs O
4. Balance H by adding H+ to side that needs H
5. Balance net charge by adding e–
6. Make e– gain equal e– loss; then add half- reactions
7. Cancel anything that is the same on both sides

25. Ion Electron Method Ex. 2 Balance in Acidic Solution
Cr2O72– + Fe2+  Cr Fe3+
1. Break into half-reactions
Cr2O72  Cr3+
Fe2+  Fe3+
2. Balance atoms other than H & O
Cr2O72  2Cr3+
Put in 2 coefficient to balance Cr
Fe already balanced

26. Ex. 2 Ion-Electron Method in Acid
3. Balance O by adding H2O to the side that needs O.
Cr2O72  2Cr3+
Right side has 7 O atoms
Left side has none
Add 7 H2O to left side
Fe2+  Fe3+
No O to balance
+ 7 H2O

27. Ex. 2 Ion-Electron Method in Acid
4. Balance H by adding H+ to side that needs H
Cr2O72  2Cr H2O
Left side has 14 H atoms
Right side has none
Add 14 H+ to right side
Fe2+  Fe3+
No H to balance
14H+ +

28. Ex. 2 Ion-Electron Method in Acid
5. Balance net charge by adding electrons.
14H+ + Cr2O72  2Cr3+ + 7H2O
6 electrons must be added to reactant side
Fe2+  Fe3+
1 electron must be added to product side
Now both half-reactions balanced for mass & charge
6e +
Net Charge =
14(+1) (–2) = 12
Net Charge =
2(+3)+7(0) = 6
+ e

29. Ex. 2 Ion-Electron Method in Acid
6. Make e– gain equal e– loss; then add half- reactions
6e + 14H+ + Cr2O72–  2Cr3+ + 7H2O
Fe2+  Fe3+ + e
7. Cancel anything that's the same on both sides
6[ ]
6e + 6Fe H+ + Cr2O72 
6Fe3+ + 2Cr H2O + 6e
6Fe H+ + Cr2O72 
6Fe3+ + 2Cr3+
+ 7H2O

30. Ion-Electron in Basic Solution
The simplest way to balance an equation in basic solution
Use steps 1-7 above, then
8. Add the same number of OH– to both sides of the equation as there are H+.
9. Combine H+ & OH– to form H2O
10. Cancel any H2O that you can from both sides

31. Ex.2 Ion-Electron Method in Base
Returning to our example of Cr2O72 & Fe2+
8. Add to both sides of equation the same number of OH– as there are H+.
9. Combine H+ and OH– to form H2O.
10. Cancel any H2O that you can
6Fe2++ 14H+
+ Cr2O72 
6Fe3+ + 2Cr3+
+ 7H2O
+ 14 OH–
+ 14 OH– 7
6Fe H2O
+ Cr2O72 
6Fe3+ + 2Cr3+
+ 7H2O + 14OH 
6Fe2+ + 7H2O
+ Cr2O72 
6Fe3+ + 2Cr3+
+ 14OH 

32. Your Turn!
Which of the following is a correctly balanced reduction half-reaction?
Fe e–  Fe°
2Fe + 6HNO3  2Fe(NO3) H2
Mn H2O  MnO4– + 8H e–
2O2–  O e–
Mg e–  Mg°

33. Ex. 3 Ion-Electron Method
Balance the following equation in basic solution:
MnO4– + HSO3–  Mn2+ + SO42
1. Break it into half-reactions
MnO4–  Mn2+
HSO3–  SO42–
2. Balance atoms other than H & O
MnO4  Mn2+
Balanced for Mn
HSO3  SO42
Balanced for S

34. Ex. 3 Ion-Electron Method
3. Add H2O to balance O
MnO4  Mn2+
HSO3  SO42
4. Add H+ to balance H
MnO4  Mn H2O
H2O + HSO3  SO42
+ 4H2O
H2O +
8H+ +
+ 3H+

35. Ex. 3 Ion-Electron Method
5. Balance net charge by adding e–.
8H+ + MnO4  Mn2+ + 4H2O
8(+1) + (–1) = = +2
Add 5 e– to reactant side
H2O + HSO3  SO42 + 3H+
0 + (–1) = –1 –2 + 3(+1) = +1
Add 2 e– to product side
5e– +
+ 2 e–

36. Ex. 3 Ion-Electron Method
6. Make e– gain equal e– loss
5e– + 8H+ + MnO4  Mn H2O
H2O + HSO3  SO42 + 3H+ + 2e–
Must multiply Mn half-reaction by 2
Must multiply S half-reaction by 5
Now have 10 e– on each side
2[ ]
5[ ]

37. Ex. 3 Ion-Electron Method
6. Then add the two half-reactions
10e– + 16H+ + 2MnO4  2Mn2+ + 8H2O
5H2O + 5HSO3  5SO42 + 15H+ + 10e–
7. Cancel anything that is the same on both sides.
Balanced in acid. 1 3
10e– + 16H+ + 2MnO4
+ 5H2O + 5HSO3
2Mn2+ + 8H2O +
5SO42 + 15H+ + 10e  
H+ + 2MnO4
+ 5HSO3
2Mn2+ + 3H2O + 5SO42 

38. Ex.3 Ion-Electron Method in Base
8. Add same number of OH– to both sides of equation as there are H+
9. Combine H+ and OH– to form H2O
10. Cancel any H2O that you can
2MnO4 + 5HSO3  2Mn2+ + 2H2O + OH SO42
H+ + 2MnO4
+ 5HSO3
2Mn2+ + 3H2O + 5SO42 
+ OH–
+ OH– 2
H2O + 2MnO4
+ 5HSO3
2Mn2+ + 3H2O + 5SO42 + OH 

39. Your Turn!
Balance each equation in Acid & Base using the Ion Electron Method.
MnO4– + C2O42–  MnO2 + CO32–
Acid: 2MnO4– + 3C2O42– + 2H2O  2MnO2 + 4H+ + 6CO32–
Base: 2MnO4– + 3C2O42– + 4OH–  2MnO2 + 2H2O + 6CO32–
ClO– + VO3–  ClO3– + V(OH)3
Acid: ClO– + 2H2O + 2VO3– + 2H+  ClO3–+ 2V(OH)3
Base: ClO– + 4H2O + 2VO3–  ClO3–+ 2V(OH)3 + 2OH–

40. Acids as Oxidizing Agents
Metals often react with acid
Form metal ions &
Molecular hydrogen gas
Molecular Equation
Zn(s) + 2HCl(aq) H2(g) + ZnCl2(aq)
Net Ionic Equation
Zn(s) + 2H+(aq) H2(g) + Zn2+(aq)
M  oxidized
H+  reduced
H+  oxidizing reagent
Zn  reducing reagent

41. Oxidation of Metals by Acids
Ease of oxidation process depends on metal
Metals that react with HCl or H2SO4
Easily oxidized by H+
More active than hydrogen (H2)
Ex. Mg, Zn, alkali metals
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
2Na(s) + 2H+(aq)  2Na+(aq) + H2(g)
Metals that don’t react with HCl or H2SO4
Not oxidized by H+
Less active than H2
Ex. Cu, Pt

42. Anion Determines Oxidizing Power
Acids are divided into 2 classes:
Nonoxidizing Acids
Anion is weaker oxidizing agent than H3O+
Only redox reaction is
2H+ + 2 e–  H2 or
2H3O+ + 2 e–  H2 + 2H2O
HCl(aq), HBr(aq), HI(aq)
H3PO4(aq)
Cold, dilute H2SO4(aq)
Most organic acids (e.g., HC2H3O2)

43. 2. Oxidizing Acids Anion is stronger oxidizing agent than H3O+
Used to react metals that are less active than H2
No H2 gas formed
HNO3(aq)
Concentrated
Dilute
Very dilute, with strong reducing agent
H2SO4(aq)
Hot, conc’d, with strong reducing agent
Hot, concentrated

44. Nitrate Ion as Oxidizing Agent
A. Concentrated HNO3
NO3– more powerful oxidizing agent than H+
NO2 is product
Partial reduction of N (+5 to +4)
NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O
Ex.
oxidation
reduction
Oxidizing agent
Reducing agent
Cu(s) + 2NO3–(aq) + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O

45. Nitrate Ion as Oxidizing Agent
B. Dilute HNO3
NO3– is more powerful oxidizing agent than H+
NO is product
Partial reduction of N (+5 to +2)
NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O
Used to react metals that are less active than H2
Ex. Reaction of copper with dilute nitric acid
3Cu(s) + 8HNO3(dil, aq)  3Cu(NO3)2(aq) + 2NO(g) + 4H2O

46. Reactions of Sulfuric Acid
A. Hot, Concentrated H2SO4
Becomes potent oxidizer
SO2 is product
Partial reduction of S (+6 to +4)
SO42– + 4H+ + 2e–  SO2(g) + 2H2O
Ex. Cu + 2H2SO4(hot, conc.)  CuSO4 + SO2 + 2H2O
B. Hot, conc’d, with strong reducing agent
H2S is product
Complete reduction of S (+6 to –2)
SO42– + 10H+ + 8e–  H2S(g) + 4H2O
Ex. 4Zn + 5H2SO4(hot, conc.)  4ZnSO4 + H2S + 4H2O

47. Your Turn!
Which of the following statements about oxidizing acids is false?
H2SO4 can behave as either an oxidizing or nonoxidizing acid, depending on the solution conditions.
Oxidizing acids can oxidize metals that are less active than hydrogen.
The anions of oxidizing acids are reduced in their reactions with metals.
Most strong acids are oxidizing acids.
Oxidizing acids are acids whose anions are stronger oxidizing agents than H+.

48. Redox Reactions of Metals
Acids reacting with metal
Special case of more general phenomena
Single Replacement Reaction
Reaction where one element replaces another
A + BC → AC + B
Metal A can replace metal B
If A is more active metal, or
Nonmetal A can replace nonmetal C
If A is more active than C
6.4 | Redox Reactions of Metals

49. Single Replacement Reaction
Left = Zn(s) + CuSO4(aq)
Center = Cu2+(aq) reduced to Cu(s); Zn(s) oxidized to Zn2+(aq)
Right = Cu(s) plated out on Zn bar
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

50. Single Replacement Reaction
Zn2+ ions take place of Cu2+ ions in solution
Cu atoms take place of Zn atoms in solid
Cu2+ oxidizes Zn° to Zn2+
Zn° reduces Cu2+ to Cu°
More active Zn° replaces less active Cu2+
Zn° is easier to oxidize!

51. Activity Series of Metals
Cu less active, can't replace Zn2+
Can't reduce Zn2+
Cu(s) + Zn2+(aq)  No reaction
General phenomenon
Element that is more easily oxidized will displace one that is less easily oxidized from its compounds
Activity Series (Table 6.3)
Metals at bottom more easily oxidized (more active) than those at top
This means that given element will be displaced from its compounds by any metal below it in table

52. How Activity Series Generated
2H+(aq) + Sr(s)  Sr2+(aq) + H2(g)
H+ oxidizes Sro to Sr2+
Sro reduces H+ to H2
More active Sro replaces less active H+
Sro is easier to oxidize!
H2 (g) + Sr2+(aq)  NO REACTION!
Why?
H2 less active, can't replace Sr2+
Can't reduce Sr2+

53. Learning Check: Metal Activity
Using the following observations, rank these metals from most reactive to least reactive:
Cu(s) + HCl(aq) → no reaction
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Mg(s) + ZnCl2(aq) → MgCl2(aq) + Zn(s)
Mg > Zn > H > Cu

54. Table 6.3 Activity Series of Some Metals

55. Reactivity Varies by Metal
M at very bottom of Table
Very strong reducing agents
Very easily oxidized
Na down to Cs
Alkali & alkaline earth metals
React with H2O as well as H+
2Na(s) + 2H2O  H2(g) + 2NaOH(aq)

56. Reactivity Varies by Metal
Ag = no reaction (top of activity series)
2HCl(aq) + Ag(s)  2AgCl(aq) + H2(g)
Zn= somewhat reactive (middle of activity series)
2HCl(aq) + Zn(s)  ZnCl2(aq) + H2(g)
Mg = very reactive (bottom of activity series)
2HCl(aq) + Mg(s)  MgCl2(aq) + H2(g)

57. Using Activity Series to Predict Reactions
If M is below H
Can displace H from solutions containing H+
2H+  H2(g)
If M is above H
Doesn't react with Nonoxidizing acids
HCl, H3PO4, etc.
In general
Metal below replaces ion above

58. Uses of Activity Series
Predictive tool for determining outcome of single replacement reactions
Given M & M'n+
Look at chart & draw arrow from M to M'n+
Arrow that points up from bottom left to top right will occur
Arrow that points down from top left to bottom right will NOT occur

59. Learning Check 2Au3+(aq) + 3Ca(s)  Au(s) + Ca2+(aq) 
Sn(s) + Na+(aq) 
Mn(s) + Co2+(aq) 
Cu(s) + H+(aq) 
2Au(s) + 3Ca2+(aq)
rxn occurs
NO reaction
NO reaction
Co(s) + Mn2+(aq)
rxn occurs
NO reaction

60. Your Turn! The activity series of metals is
Au < Ag < Cu < Sn < Cd < Zn < Al < Mg < Na < Cs
(least active) (most active)
Based on this list, which element would undergo reduction most readily? Ag Al Cu Cd Zn

61. Oxygen as an Oxidizing Agent
Oxygen Reacts With Many Substances
Combustion
Rapid reaction of substance with oxygen that gives off both heat and light
Hydrocarbons are important fuels
Products depend on how much O2 is available
1. Complete Combustion
O2 plentiful
CO2 & H2O products
Ex. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O
6.5 | Molecular Oxygen as an Oxidizing Agent

62. Oxidation of Organic Compounds
2. Incomplete Combustion
Not enough O2
a. Limited O2 supply
CO is carbon product
2CH4(g) + 3O2(g)  2CO(g) + 4H2O
b. Very limited O2
C(s) is carbon product
CH4(g) + O2(g)  C(s) + 2H2O
Gives tiny black particles
Soot—lamp black
Component of air pollution

63. Oxidation of Organic Compounds
3. Combustion of Organics containing O
Still produce CO2 & H2O
Need less added O2
C12H22O11(s) + 12 O2(g)  12 CO2(g) H2O
4. Combustion of Organics containing S
Produce SO2 as product
2C4H9SH + 15O2(g)  8CO2(g) + 10H2O + 2SO2(g)
SO2 turns into acid rain when mixed with water
SO2 oxidized to SO3
SO3 reacts with H2O to form H2SO4

64. B. Reaction of Metals with O2
Corrosion
Direct reaction of metals with O2
Many metals corrode or tarnish when exposed to O2
Ex.
2Mg(s) + O2(g)  2MgO(s)
4Al(s) + 3O2(g)  2Al2O3(s)
4Fe(s) + 3O2(g)  2Fe2O3(s)
4Ag(s) + O2(g)  2Ag2O(s)

65. C. Reaction of Nonmetals with O2
Many nonmetals react directly with O2 to form nonmetal oxides
Sulfur reacts with O2
Forms SO2
S(s) + O2(g)  2SO2(g)
Nitrogen reacts with O2
Forms various oxides
NO, NO2, N2O, N2O3, N2O4, and N2O5
Dinitrogen oxide, N2O
Laughing gas used by dentists
Propellant in canned whipped cream

66. Learning Check: Complete Following Reactions
Aluminum metal and oxygen gas forms aluminum oxide solid
Solid sulfur (S8) burns in oxygen gas to make gaseous sulfur trioxide
Copper metal is heated in oxygen to form black copper(II) oxide solid
4Al(s) + 3O2(g) → 2Al2O3(s)
S8(s) + 12O2(g) → 8SO3(g)
All ionic compounds are solids at room temperature.
2Cu(s) + O2(g) → 2CuO(s)

67. Your Turn! Which of the following reactions is not a redox reaction?
Na2S(aq) + MnCl2(aq)  2NaCl(aq) + MnS(s)
CH4(g) + O2(g)  C(s) + 2H2O
2Zn(s) + O2(g)  2ZnO(s)
Cu(s) + 4H+(aq) + 2NO3–(aq)  Cu2+(aq) + 2NO2(g) H2O
Sr(s) + 2H+(aq)  Sr2+(aq) + H2(g)

68. Stoichiometry in Redox Reactions
Like any other stoichiometry problem
Balance redox reaction
Use stoichiometric coefficients to relate mole of 1 substance to moles of another
Types of problems
Start with mass or volume of one reactant & find mass or volume of product
Perform titrations
Have limiting reactant calculations
Calculate % yields
6.6 | Stoichiometry of Redox Reactions

69. Stoichiometry in Redox Reactions
Ex. How many grams of Na2SO3 (126.1 g/mol) are needed to completely react with 12.4 g of K2Cr2O7 (294.2 g/mol)?
1st need balanced redox equation
8H+(aq) + Cr2O72(aq) + 3SO32(aq)  3SO42(aq)
+ 2Cr3+(aq) + 4H2O
Then do calculations
1. g K2Cr2O7  moles K2Cr2O7  moles Cr2O72(aq)
2. moles Cr2O72(aq)  moles 3SO42(aq)
3. moles SO32(aq)  moles Na2SO3  g Na2SO3

70. Stoichiometry Example (cont)
grams K2Cr2O7  moles K2Cr2O7  moles Cr2O72 (aq)
moles Cr2O72 (aq)  moles 3SO32 (aq)
moles SO32 (aq)  moles Na2SO3  g Na2SO3

71. Redox Titrations
Equivalence point reached when # of moles of oxidizing & reducing agents have been mixed in the correct stoichiometric ratio
No simple indicators to detect endpoints
3 very useful oxidizing agents that change color
1. KMnO4: Deep purple of MnO4 fades to almost colorless Mn2+ (very pale pink)
2. K2Cr2O7: Bright yellow orange of Cr2O72 changes to pale blue green of Cr3+
3. IO3 : When reduced to I2(s) in presence of I, forms I3 which forms dark blue complex with starch

72. Redox Titration Example
I reacts with IO3 in acidic solution to form I2(s). If mL of M I is needed to titrate mL of a solution containing IO3, what is the M of the solution?
1. Write Unbalanced Equation 
I(aq) + IO3(aq)  I2(s)
I(aq) is oxidized to I2
IO3(aq) is reduced to I2

73. Redox Titration Example (cont)
2. Balance Equation
Note: we are in acidic solution
2I(aq)  I2(s) + 2e
Not done as not lowest whole number coefficients
5I(aq) + IO3(aq) + 6H+(aq)  3I2(s) + 3H2O
5  [ ]
2IO3(aq) + 12H+(aq) + 10e  I2(s) + 6H2O
10I(aq) + 2IO3(aq) + 12H+(aq)  6I2(s) + 6H2O 2 2 2 2 2

74. 3. Now for the Calculations
Calculate mmol of I– titrated
Convert to mmol of IO3– present
Convert to M of IO3– solution
= M IO3–

75. Ore Analysis M of KMnO4  V = mol KMnO4
A g sample of tin ore was dissolved in acid solution converting all the tin to tin(II). In a titration, mL of M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample?
M of KMnO4  V = mol KMnO4
mol KMnO4  mol Sn/mol KMnO4 = mol Sn2+
mol Sn2+  MM = g Sn2+ in sample
%Sn = g Sn/g sample  100 %
3Sn2+(aq) + 2MnO4(aq) + 8H+(aq)  Sn4+(aq) + 2MnO2(s) + 4H2O

76. Tin Ore Analysis Continued
M of KMnO4  V = mmol KMnO M KMnO4  8.08 mL = mmol KMnO4 mmol KMnO4  mmol MnO4  mmol Sn2+ Mol Sn2+  g/mol = g Sn in original sample %Sn = g Sn/ g sample  100 %
= mmol Sn2+
= g Sn
= 23.97% Sn

77. Your Turn!
The amount of hydrogen peroxide (H2O2, MM = g/mol) in hair bleach was determined by titration with a standard KMnO4 (MM = g/mol) solution:
2MnO4–(aq) + 5H2O2(aq) + 6H+(aq)  5O2(g) + 2Mn2+(aq) + 8H2O
If 43.2 mL of M MnO4– was needed to reach the endpoint, how many grams of H2O2 are in the sample of hair bleach?
0.771 g
0.386 g g
386 g
154 g
= g H2O2